Repeat Problem 3.84 when each of the 3 players

selects from his own urn. That is, suppose that there are

3 different urns of 12 balls with 4 white balls in each urn.

Given information:

A, B, C each have their own charnel containing 12 balls out of which 4 are white. A, B, C draw one ball from the charnel in race. The earliest one to draw a white ball wins.

Formula used:

Probability of an event$=\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}$.

Sum of chances of all possible issues is outcomes is $1$

Calculation:

If the balls are drawn with replacement, A can win in the first turn with probability $\frac{4}{12}=\frac{1}{3}$

If A doesn't win in the first turn with probability $1-\frac{1}{3}$, Also B and C should both withdraw non-white balls with probability $\frac{2}{3}$. Also A can win in the alternate turn with the probability role="math" localid="1646722256960" $\frac{1}{3}·{\left[\frac{2}{3}\right]}^3$.

Also , the chances of A winning in the $i^{th}$turn is $\frac{1}{3}·{\left[\frac{2}{3}\right]}^{3\left(i-1\right)}$

Hence, probability of A winning is $\frac{1}{3}\sum _{i=1}^{\infty }{\left[\frac{2}{3}\right]}^{3(i-1)}=\frac{9}{19}$.

Also , B can win in the first turn with the probability $\frac{2}{3}·\frac{1}{3}$if A loses in the first turn.

B can win in the alternate turn if A, B, and C lose in the first turn all with probability $\frac{2}{3}$and A loses in the alternate turn again with the same probability. Hence, B can win in the alternate turn with the probability $\frac{1}{3}·{\left[\frac{2}{3}\right]}^4$.

B can win with probability $\frac{1}{3}·\frac{2}{3}\sum _{i=1}^{\infty }{\left[\frac{2}{3}\right]}^{3(i-1)}=\frac{2}{9}\frac{1}{19}=\frac{6}{19}$.

C can with probability $1-\frac{9}{19}-\frac{6}{19}=\frac{4}{19}$.

Given information:

$\mathrm{A},\mathrm{B},\mathrm{C}$each have their own charnel containing 12 balls out of which 4 are white. $\mathrm{A},\mathrm{B},\mathrm{C}$draw one ball from the charnel in race. The first one to draw a white ball wins.

Formula used:

- Probability of an event $=\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}$

- Sum of chances of all possible issues is outcomes is $1$.

Calculation:

If the balls are not replaced, $A$can win in the first turn with the probability $\frac{1}{3}$

If $A$does not win in the first turn, $B$and $C$must also draw non-white balls. After A has drawn a non-white ball in the first turn, B has 8 non-white balls to draw from and $C$has 8 non-white balls to draw from. Hence, A can win in the alternate turn with the probability ${\left[\frac{8}{12}\right]}^3\frac{4}{11}$

A can win in the third turn also with a probability ${\left(\frac{8}{12}\right)}^3{\left(\frac{7}{11}\right)}^3\frac{4}{10}$.

Hence, A can win in any of the turns before $9^{\text{th}}$turn since $9^{\text{th}}$turn will be a definite palm for A.

The sum of all chances for $\mathrm{A}$to win $=0.3884$.

B can win in the first turn if A draws a non-white ball. The probability of B winning is $\frac{8}{12}·\frac{4}{12}$.

B can win in the second turn if all three players draw non-white balls with probability ${\left(\frac{8}{12}\right)}^3\left(\frac{7}{11}\right)\frac{4}{11}$.

B can win in the third turn if all three players draw non-white balls with probability ${\left(\frac{8}{12}\right)}^3{\left(\frac{7}{11}\right)}^3\frac{6}{10}\frac{4}{10}$.

B has to win before or on the $8^{\text{th}}$turn since A will surely win on the $9^{\text{th}}$turn.

Probability of $B$winning is sum of winning in all turns $=0.3138$

Probability of $C$winning is $1-0.3884-0.3138=0.2978$